A) `CCl_(4)` does not act as lewis acid
B) Silanes are strong reducing agents
C) Crystalline silica will have Diamond like structure
The correct answer is

C Cl_4 does not act as a Lewis acid while SiCl_4 and SnCl_4 act as lewis acids why?

C Cl_(4) does not act a Lewis acid while SiCl_(4) behaves as a Lewis acid. Discuss.

C Cl_(4) does not act as Lewis acid while SiCl_(4) and SnCl_(4) act as Lewis acids. Why?

A) Silanes are good reducing agents B) SiO_(2) is a giant tetrahedral polymer C) SnCl_(4) act as Bronsted Base

Alkali and alkaline earth metals have low ionisation enthalpies and hence exhibit characteristic flame colouration. They have high negative electrode potentials and hence are strong reducing agents. They dissolve in liquid ammonia to give a solution which conducts electricity and act as strong reducing agent. being stronger reducing agent than hydrogen, they are usually prepared by the electrolysis of their fused chlorides. Their oxides are basic and the basic strength increases down the group. The solubility of carbonates and sulphates of alkali and alkaline earth metals show opposite trends. only the carbonates of Li and alkaline earth metals decompose on heating. The bicarbonates of both alkali and alkaline earth metals on heating give carbonates. The compound insoluble in acetic acid is

Carbon oxygen double bond are easily reduced by NaBH_(4) or LiAlH_(4) . The actual reducing agent in these reduction is hrdride ion (H^(-)) The metal hydrogen bond in LiAlH_(4) is more than polar than metal hydrogen bond in NaBH_(4) . As a result LiAlH_(4) is strong reducing agent than NaBH_(4) . Esters, carboxylic acids, amides cannot be reduced by NaBH_(4) The carbonyl group of amide of reduced to methylene group by LiAlH_(4) Find the correct product of the following reaction:

Carbon-oxygen double bond are easily reduced by NaBH_(4) or LiAlH_(4) . The actual reducing agent in these reduction is hydride ion (H^(-)) . The metal-hydrogen bond in LiAlH_(4) is more polar than metal-hydrogen bond in NaBH_(4) . As a result LiAlH_(4) is strong reducing agent than NaBH_(4) . Esters, carboxylic acids, amides cannot be reducted by NaBH_(4) . The carbonyl group of amide to methylene group by LiAlH_(4) . Find the correct product of the following reaction :